Question

The electron concentration in an $n$-type semiconductor is the same as hole concentration in a $p$-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

• A. current in n-type = current in p-type
• B. current in p-type > current in n-type
• C. current in n-type > current in p-type
• D. No current will flow in p-type, current will only flow in n-type

Answer 1

The Correct Option is C

To answer this question, we need to consider the conductivity and mobility of electrons and holes in semiconductors.

1. In both \(n\)-type and \(p\)-type semiconductors, the current (\(I\)) due to an applied electric field can be described by the equation: \(I = nq\mu E A\), where:
   • \(n\): Carrier concentration (number of charge carriers, either electrons or holes)
   • \(q\): Charge of the carrier (for electrons, it's \(-1.6 \times 10^{-19}\,\text{C}\); for holes, it's \(1.6 \times 10^{-19}\,\text{C}\))
   • \(\mu\): Mobility of the charge carrier
   • \(E\): Electric field
   • \(A\): Cross-sectional area through which the current flows
2. In an \(n\)-type semiconductor, the majority carriers are electrons, while in a \(p\)-type semiconductor, the majority carriers are holes.
3. We are given that the electron concentration in the \(n\)-type semiconductor is equal to the hole concentration in the \(p\)-type semiconductor.
4. The mobility of electrons (\(\mu_e\)) is generally higher than the mobility of holes (\(\mu_h\)) in semiconductors. This means: \(\mu_e > \mu_h\).
5. With the same carrier concentration and electric field, the current in the \(n\)-type semiconductor will be proportional to \(\mu_e\), while the current in the \(p\)-type semiconductor will be proportional to \(\mu_h\).

Since \(\mu_e > \mu_h\), the current in the \(n\)-type semiconductor will be greater than the current in the \(p\)-type semiconductor.

Therefore, the correct answer is: current in n-type > current in p-type.